TITLE: A collection of sets that cover all edges in Kn?
QUESTION [5 upvotes]: The problem is the following:
Let $\mathcal{F}$ be a family of distinct proper subsets of {1,2,...,n}. Suppose that for every $1\leq i\neq j\leq n$ there is a unique member of $\mathcal{F}$ that contains both $i$ and $j$. Prove that $\mathcal{F}$ has at least $n$ elements.
I realized that each member $A_i$ in $\mathcal{F}$ can be regarded as a collection of edges in a clique induced by the element in $A_i$, and at the end all the members in $\mathcal{F}$ exactly cover all the edges in $K_n$ and every edge must live in exactly one member in $\mathcal{F}$. So if we write $\mathcal{F}=\{A_1,...,A_m\}$ with sizes $a_1,...,a_m$, respectively, then we obtain the following equation:
$a_1\choose 2$+...+$a_m\choose 2$ = $n\choose 2$ =$\frac{n(n-1)}{2}$.
I got stuck from here. Even if I assumed $m<n$, I cannot obtain a contradiction. Is my approach correct at all? If so, how should I continue? I appreciate any help and insights.

REPLY [3 votes]: This result is known as the de Bruijn–Erdős theorem. It is usually stated in terms of incidence geometries. An incidence geometry has:

A set of objects called points.
A collection of sets of points called lines. If a line $\ell$ contains a point $P$, we also say that $P$ lies on $\ell$, or $\ell$ passes through $P$.

We require that for any two points, there is a unique line passing through both.
We also make some nontriviality assumptions. Each line contains two points (though this doesn't matter for this question, since single-point lines only help us). There is not a line containing all the points (which does matter for your question: the subsets in $\mathcal F$ must be proper subsets, or the conclusion is false).
In your case, $\{1,2,\dots,n\}$ is the set of points and $\mathcal F$ is the set of lines.

The de Bruijn–Erdős theorem simply says that in any incidence geometry, there are at least as many lines as points.
If you've gone ahead and read the Wikipedia article I linked to earlier, then you've seen the proof there, which assumes a Euclidean incidence geometry: one in which the points are some finite set of points in the plane, and the lines are actual lines drawn through those points. This is not the proof that de Bruijn and Erdős gave, for which you can read their original paper; their proof applies to any abstract incidence geometry, and so it solves your problem.
A few years ago, I also wrote an alternative writeup of their proof, which you can find here.